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Let $\Gamma$ be a simplicial graph and $\widehat{\Gamma}$ be the corresponding clique complex (the flag complex obtained after adding simplices for each compete graph). We can costruct the right-angled Artin group $A_\Gamma$ and consider the outer automorphism group $\mathrm{Out}(A_\Gamma)$. It is well known that $\mathrm{Out}(A_\Gamma)$ is a finite group iff there is no $v\in\Gamma$ such that $\Gamma\setminus\mathrm{st}(v)$ is disconnected and there are no $u,v\in\Gamma$ with $u\neq v$ and $\mathrm{lk}(u)\subset\mathrm{st}(v)$.

I claim that $\widehat{\Gamma}$ simply connected implies that $\mathrm{Out}(A_\Gamma)$ is infinite. What my intuition tells me is that $\widehat{\Gamma}$ being simply connected removes the possiblity of induced cylces of length $n\geq4$ while $\mathrm{lk}(u)\subset\mathrm{st}(v)$ can happen if $d(u,v)\leq 2$ and they don't lie in a induced cylces of length $n\geq5$. So it is like one of the conditions is contrary to the other.

I know this is not a proof and that there might be some weird counterexamples to this result, but I feel that this should be true. Thanks for your help.

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What about the following tesselation of the sphere? Am I missing something if I say that there are no disconnecting stars nor two distinct vertices such that the link of one is contained in the star of the other?

enter image description here

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  • $\begingroup$ I was thinking about something that looks trivially true but I have no idea how to prove. In the same spirit of this example there must exist a triangulation of the $n$-sphere with no disconnecting stars nor two distinct vertices such that the link of one is contained in the star of the other, right? (Thanks for the correction) $\endgroup$
    – Marcos
    Commented Jun 5 at 8:00
  • $\begingroup$ I think so. And probably not only for spheres, but also for arbitrary manifolds (with respect to a sufficiently good triangulation). $\endgroup$
    – AGenevois
    Commented Jun 6 at 4:44

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